Isaac Sarver
1) You have been hired as a consultant for ”death defying” stunts. Your task is to design a stunt
in which a 686N(70kg) actor who is 1.7 m tall jumps off the top of a 30m tall tower with an elastic
bungee cord tied to one ankle with the other end of the cord tied to the top of the tower. The 22m
long cord is very light (relatively massless) but very strong and needs to stretch so that it can stop
the actor without pulling a leg off. Such a cord acts like a spring. For maximum dramatic effect,
the jump will be off a diving board at the top of the tower where the person first jumps straight
vertically upward at a speed of 3 m/s, reaches a maximum height and then falls straight vertically
downward, head pointing down. To minimize the force that the cord exerts on the leg, you want it
to stretch as far as possible but only to the extent that the head barely touches the ground (else
the person will die!) after which the elastic spring-like cord will begin lifting the person back up.
A) Given that you know that the length of the cord is 22m and that it is relatively massless, what
other physical characteristic/property of the cord should you know in order to ensure the safety of
this stunt? Estimate a numerical value for that quantity so that you can purchase a suitable 22m
long cord. You may ignore the effects of air resistance.
B) Where along the entire distance of the fall is the magnitude of the acceleration of the person
(and hence the net force) a maximum and where is it a minimum. Give numerical values for both
the location and the values for the maximum and minimum acceleration.
I’m going to use energy conservation to solve the problem. Since the kinetic energy is dependent on a v 2 , the stunt man’s jump up at 3 ms is the same him jumping down at the same speed. So
I’m just going to invert the man at the same height and send him down at the speed given.
686N
m 2
) = 70kg9.8 sm2 (30m − 1.7m) + 12 70kg(3 ms )2
Ei = P Ei + KEi = 686N (30m − 1.7m) + 21 9.8
m (3
s
2
s
Ei = 19.7kJ
Ef = P Ef = 12 k(30m − 22m − 1.7m)2 = 19.845m2 k
N
19.7kJ
Ei = Ef ⇒ k = 19.845m
2 = 994.1 m
Now I need to find the extrema of acceleration. Since acceleration is proportional to force, I’ll look
for the extrema of the force applied.
I know where to find the negative extrema of acceleration, and that is while the man is in free fall
from the place that he jumps up to the place that the bungee catches him 22m below his starting
point and with his feet 8 meters in the air. That acceleration is −9.8 sm2
I know where to find the positive extrema and that is at the end of his jump with his hair dusting
the ground.
F = mg − k(8m − 1.7m) = −686N − 994.1 N
(−6.3m) = 5577N
m
F
5577N
m
a = m = 70k = 79.7 s2 = 8.13g
The absolute minimum acceleration is when the force is 0N.
mg − kx = 0N
686N
x = mg
= 994.1
N = .6900m
k
m
The acceleration of 0 sm2 happens when the bungee stretches .69m, or when his feet are 22.69m below
the platform, or when his head has 5.61m to go.
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2) Use conservation of momentum (d~p = d(m~v ) = 0N s) to derive the rocket equation. Essentially,
determine how the change in velocity of a rocket depends on the exhaust velocity, initial mass, and
final mass using the fact that the rocket+exhaust system must conserve momentum.
Find the final mass of a VASIMR as percent of the initial mass for a trans-lunar injection maneuver.
∆m
= .1 gs , ∆v = 3.05 km
, ve = 50 km
(Mach 146.9)
∆t
s
s
∆m
Repeat for the J-2 rocket. ∆t = 221.6 kg
, ve = 4.512 km
(Mach 13.3)
s
s
Why do you think we use chemical rockets, like the J-2, for launches and human space travel instead
of ion engines like we do for space probes?
Mi
Mf
= exp(− ∆v
)
ve
F = ma = ve ∆m
∆t
VASIMR (Ion Rocket) :
Derivation first:
d~p = d(m~v ) = ~v dm + md~v = 0N s
~v dm R= −md~v R
~v
M
= ~vif d~v
− ~ve Mif dm
m
exp(−
3.05 km
s
) = 94.08%
50 km
s
.0001 kg
50000 ms
s
M
− ~ve ln(m)|Mfi = ∆~v
M
~ve ln( Mfi ) = ∆~v
F =
= 5N
J − 2 (Chemical Rocket) :
3.05 km
s
) = 50.87%
Now I need to rearrange the equation and apply exp(− 4.512 km
s
the numbers given.
4512 ms = 1000000N
F = 221.6 kg
s
Mf
~ve ln( Mi ) = ∆~v
As you can see, the ion rocket is much more efficient than the chemical rocket. The problem that it
has is that doesn’t have the force needed to launch things or make them move in any particularly
reasonable amount of time. Space is highly irradiated and we doen’t want to leave humans in it
any longer than necessary. So we use chemical rockets when we want to move now. Ion rockets are
used on space probes because we have the time to wait for the maneuver to complete. I hear that
there is an ion rocket is going to the space station for testing and for orbital boosts that it needs
due to atmospheric drag. There again, we have time to wait for the maneuver to complete.
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3) Consider the Cornell Potential that is used for quarkonia, U (r) = −.09866GeV · f m 1r + 1 GeV
r.
fm
If I tried to pull the bottom quark away from the anti-bottom quark, at what point would an
up/anti-up pair of quarks pop into existence? Repeat for a bottom/anti-bottom pair. If the proton
and is made of two up and a down quark, where does most of your mass
has a mass of 938.3 MceV
2
M eV
, mu = 2.2 MceV
I’ll tell you why we go on and on about the
come from? mb = 4.2 GeV
2 , md = 5 c2
c2
Higgs boson producing mass.
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This potential is not as bad as could have been. When I got it, I saw U (r) = − 2r
+ .1973GeV 2 r
because h̄ and c where both made to be 1 in appropriate units.
So the idea here is to find out how much mass the quark pairs are and then to figure out where the
potential has that much energy in it.
= 4.4 MceV
2md = 2 × 2.2 MceV
2
2
1
− .09866GeV · f m r + 1 GeV
r = 4.4M eV
fm
1
eV
− 98.66M eV · f m r − 4.4M eV + 1000 Mf m
r = 0M eV
Because I have zero on the left hand side, I can multiply by MreV with no effect.
−98.66f m − 4.4r + 1000 f1m r2 = 0f m
r=
q
4.4+ 4.42 −4(1000 f1m )(−98.66f m)
2000 f1m
r = .316f m
= 8.4 GeV
2mb = 2 × 4.2 GeV
c2
c2
− .09866GeV · f m 1r + 1 GeV
r = 8.4GeV
fm
1 2
− .09866f m − 8.4r + 1 f m r = 0f m
r=
q
8.4+ 8.42 −4(1 f1m )(−.09866f m)
2 f1m
r = 8.41f m
+ 5 MceV
= 9.4 MceV
2 ∗ mu + md = 2 × 2.2 MceV
2
2
2
M eV
The remaining 928.9 c2 of the proton mass is missing. That mass comes from a complicated phenomena that comes from QCD. It comes from the same place that the 1 GeV
r part of the potential
fm
came from. It isn’t exactly the same for protons. Essentially, the energy stored in the potential
becomes mass of the whole particle. The reason we carry on about the Higgs boson giving mass to
quarks and such is because without that mass given to them from the Higgs mechanism there no
bonding energy. And without bonding energy, there are no bonds. Without bonds, we fall apart.
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4) Little brother was cruising down Idaho Highway 16 at 55mph in his Toyota Corolla. An intoxicated man crosses Beacon Light Rd in his Porsche 911 at 10 mph. (1mph = .447 ms ) The Corolla
has a mass of 1300 kg. The Porsche weighs in 1400kg. Little brother T-bones the man. The door of
the Porsche collapses 35 cm. The bumper of the Corolla compresses into the engine compartment
10 cm. The collision is mostly elastic. How fast are the cars moving after the collision. What kind
of acceleration did the two experience. If 100g is lethal, did either survive? Why do we have air
bags and crumple zones in cars?
So in this case energy and momentum are conserved enough to use the equations. I also happen to have it that each car is initially traveling at right angles to each other. That means that in
the x and y axis I can treat it as a moving object running into a stationary object. This allows me
to show the equations once and then use the results again without having to go through the whole
proof again.
p~i = p~f
m2
v2f
m1 v1i = m1 v1f + m2 v2f ⇒ v1i = v1f + m
1
Ei = Ef
1
2
2
+ 21 m2 v2f
m v 2 = 12 m1 v1f
2 1 1i
m2
2
2
2
m1 (v1f + m1 v2f ) = m1 v1f + m2 v2f
m2
m2
2
2
2
2
m1 (v1f
+ 2m
) = m1 v1f
+ m2 v2f
v1f v2f + m22 v2f
1
1
m2
2v1f + m
v2f = v2f In this line I divide out an extra v2f which will only produce a v2f of zero meters
1
per second which is the initial condition. Notice that because energy is conserved, the collision is
time reverse invariant, meaning it will go forward just as well as it will go backward. Removing
that extra v2f removes the initial conditions as a solution.
m2
)
2v1f = v2f (1 − m
1
m1 −m2
v1f = 2m1 v2f
1 −m2
m1 v1i = m1 m2m
v2f + m2 v2f
1
m1 +m2
m1 v1i = 2 v2f
2m1
v = v2f
m1 +m2 1i
m2 2m1
2
v1i = v1f + m
v1i = v1f + m2m
v1i
1 m1 +m2
1 +m2
m1 +m2
2m2
v − m1 +m2 v1i = v1f
m1 +m2 1i
m1 −m2
v = v1f
m1 +m2 1i
Now that I have the results for one object running into a stationary second object, I can apply to
the 2-D case of two objects hitting each other at right angles.
m1 −m2
2
~v1f = ( m
v1i , m2m
v2i )
1 +m2
1 +m2
2m1
m2 −m1
~v2f = ( m1 +m2 v1i , m1 +m2 v2i )
m/s
1300kg−1400kg
2×1400kg
~vCf = ( 1300kg+1400kg
55mph, 1300kg+1400kg
10mph) × 2.24mph
= (−.911, 4.64) ms
2×1300kg
~vP f = ( 1300kg+1400kg
55mph, 1400kg−1300kg
10mph) ×
1300kg+1400kg
m/s
2.24mph
= (23.7, .166) ms
At this point we enter in to into a fuzzy zone. I only have enough information to assume constant
acceleration over the displacement and duration of the collision. I can find acceleration in the direction of the Corrolla’s travel and then I can find the duration of the collision. From that I can
find the acceleration in the direction of the Porsche’s travel. I say this is going to be fuzzy on the
account I don’t think I’m guaranteed that the time of displacement from the two cars will be equal.
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v −v
vf2 = vi2 + 2ad
tP = f a 0 = .0380s
Guess I had nothing to be worried about.
v −v
a = ft i
4.64 m
−0 m
s
s
aCP = .0380s
= 121.95 sm2
.166 m
−4.4704 m
s
s
aP P =
= −113.25 sm2
.0380s
aC = (−670.78, 121.95) sm2 = 681 sm2 = 69.52g
aP = (622.87, −113.25) sm2 = 633 sm2 = 64.56g
vf2 −vi2
2d
(−.911 m
)2 −(24.5872 m
)2
s
s
= −670.78 sm2
aCC =
2(.45m)
(−23.7 m
)2 −(0 m
)2
s
s
= 622.87 sm2
aP C =
2(.45m)
v −v
vf = v0 + at ⇒ t = f a 0
m
m
−.911 s −24.5872 s
= .0380s
tC =
−670.78
a=
According to this both live with injury. I know that in this particular instance, the drunk man died
and little brother lived to tell the tale. This suggests one three possibilities in this case. 1) Average
acceleration is not constant and the drunk man peaked at a higher g-load especially seeing as how
he was nailed in the driver-side door. 2) My numbers aren’t as good as I should like. 3) Or we
don’t know as much about physiology as we think we do. Essentially I don’t know as much as I
should like. Especially as little brother’s car ended up on the east curb and not the west curb.
The reason for the crumble zones and other such safety features is to make the acceleration of the
passengers is as low as possible to ensure that they take as little damage as possible.
5) Texas A&M played Auburn this weekend. (I picked Mr. Ennis because I thought he was a
good choice. As it happens, he was benched with a torn ACL.) #42 Kirby Ennis sacked #14 Nick
Marshall of the Auburn Tigers. Mr. Ennis had a mass of 140kg and Mr. Marshall had a mass of
95kg. Mr. Ennis was running full bore at Mr. Marshall at 5.2 ms . Mr. Marshall was in full retreat
attempting to get the ball off before he was sacked. So he was retreating at 3.3 ms . They collide
completely inelastically. How fast were they moving after the collision? Where does the lost energy
go?
This is a case of momentum conservation.
mE = 140kg, mM = 95kg
vE = 5.2 ms , vM = 3.3 ms
pi = pf
mE vE + mM vM = (mE + mM )vf
mE vE +mM vM
= vf
mE +mM
vf =
140kg5.2 m
+95kg3.3 m
s
s
140kg+95kg
= 4.43 ms
Now I want to find the energy loss.
KE = 12 mv 2
KEE = 21 140kg(5.2 ms )2 = 1892.8J
KEM = 12 95kg(3.3 ms )2 = 517.275J
KET = KEE + KEM = 2410.075J
KEf = 12 (95kg + 140kg)(4.43 ms )2 = 2307.9J
KET − KEf = 102.2J
That 102J of energy went into the deformation of the two players and into the generation of
sound. That energy was converted to heat via intermolecular friction in whatever that got deformed.
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